Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int r, c;
  cin >> r >> c;
  int a[r][c];
  int ctoz = 0;
  bool poss = 1;
  if (r < c) {
    int m[r];
    for (int i = 0; i < r; i++) {
      int min = 600;
      for (int j = 0; j < c; j++) {
        cin >> a[i][j];
        if (min > a[i][j]) min = a[i][j];
      }
      m[i] = min;
      ctoz += min;
    }
    for (int zz = 0; zz < c; zz++) {
      ctoz += a[0][zz] - m[0];
    }
    for (int t = 0; t < c; t++) {
      for (int y = 0; y < r - 1; y++) {
        if (a[y][t] - m[y] != a[y + 1][t] - m[y + 1]) poss = 0;
      }
    }
    if (poss == 0)
      cout << "-1" << endl;
    else {
      cout << ctoz << endl;
      for (int u = 0; u < r; u++) {
        for (int h = 0; h < m[u]; h++) {
          cout << "row " << u + 1 << endl;
        }
      }
      for (int v = 0; v < c; v++) {
        while (a[0][v] > m[0]) {
          cout << "col " << v + 1 << endl;
          a[0][v]--;
        }
      }
    }
  } else {
    int m[c];
    for (int i = 0; i < r; i++) {
      for (int j = 0; j < c; j++) cin >> a[i][j];
    }
    for (int i = 0; i < c; i++) {
      int min = 600;
      for (int j = 0; j < r; j++) {
        if (min > a[j][i]) min = a[j][i];
      }
      m[i] = min;
      ctoz += min;
    }
    for (int zz = 0; zz < r; zz++) {
      ctoz += a[zz][0] - m[0];
    }
    for (int t = 0; t < r; t++) {
      for (int y = 0; y < c - 1; y++) {
        if (a[t][y] - m[y] != a[t][y + 1] - m[y + 1]) {
          poss = 0;
        }
      }
    }
    if (poss == 0)
      cout << "-1" << endl;
    else {
      cout << ctoz << endl;
      for (int u = 0; u < c; u++) {
        for (int h = 0; h < m[u]; h++) {
          cout << "col " << u + 1 << endl;
        }
      }
      for (int v = 0; v < r; v++) {
        while (a[v][0] > m[0]) {
          cout << "row " << v + 1 << endl;
          a[v][0]--;
        }
      }
    }
  }
}